Proving $n^3$ is even iff $n$ is even

HINT: For the second implication, try proving the contrapositive of the implication. Suppose $n$ is not even (i.e., assume $n$ is odd), and prove that, then, $n^3$ is not even (i.e., $n^3$ is odd).

$$P \implies Q \equiv \lnot Q \implies \lnot P$$

Note that $n^3-n=n(n-1)(n+1)$. This is always even. Since the a difference $x-y$ is even $\iff$ both $x,y$ are even or both $x,y$ are odd, you're done.

Although I agree that the other comment and answer given are correct and are likely the approach you are expected to use, personally I immediately note that if $n$ is written in its prime factorization form ($p_1^{e_1} \cdot p_2^{e_2} ... p_i^{e_i}$) and you observe the form that $n^3$ has as well, the stated proposition is obvious. But, I probably only see that so easily because I am obsessed with the prime factorization patterns of numbers.

Note that if you are taking a proof-writing class, this is a perfect example of using one proof as necessary for another, since the fact that no new factors are introduced depends on the Fundamental Theorem of Arithmetic.